A053820 a(n) = Sum_{k=1..n, gcd(n,k) = 1} k^4.

1, 1, 17, 82, 354, 626, 2275, 3108, 7395, 9044, 25333, 17668, 60710, 50470, 88388, 103496, 243848, 129750, 432345, 266088, 497574, 497178, 1151403, 539912, 1541770, 1153724, 1900089, 1516844, 3756718, 1246568, 5273999

Offset: 1

Author

N. J. A. Sloane, Apr 07 2000

Keywords

• nonn

Comments

If gcd(n,30) = 1, then a(n) is divisible by n. If n has at least one prime factor == 1 (mod 30), then a(n) is divisible by n. - Jianing Song, Jul 13 2018

References

• Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 48, problem 15, the function phi_4(n).
• L. E. Dickson, History of the Theory of Numbers, Vol. I (Reprint 1966), p. 140.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
• John D. Baum, A Number-Theoretic Sum, Mathematics Magazine 55.2 (1982): 111-113.
• P. G. Brown, Some comments on inverse arithmetic functions, Math. Gaz. 89 (2005) 403-408.

Crossrefs

Column k=4 of A308477.

Cf. A000010, A023900, A053818, A053819, A063453.

Programs

Mathematica

a[n_] := Sum[If[GCD[n, k] == 1, k^4, 0], {k, 1, n}]; Table[a[n], {n, 1, 31}] (* Jean-François Alcover, Feb 26 2014 *)
a[1] = 1; a[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; (n^4/5) * Times @@ ((p - 1)*p^(e - 1)) + (n^3/3) * Times @@ (1 - p) - (n/30) * Times @@ (1 - p^3)]; Array[a, 100] (* Amiram Eldar, Dec 03 2023 *)

PARI

a(n) = sum(k=1, n, (gcd(n,k) == 1)*k^4); \\ Michel Marcus, Feb 26 2014

PARI

a(n) = {my(f = factor(n)); if(n == 1, 1, (n^4/5) * eulerphi(f) + (n^3/3) * prod(i = 1, #f~, 1 - f[i, 1]) - (n/30) * prod(i = 1, #f~, 1 - f[i, 1]^3));} \\ Amiram Eldar, Dec 03 2023

Formula

a(n) = (6*n^4*A000010(n)+10*n^3*A023900(n)-n*A063453(n))/30 for n>1. Formula is derived from a more general formula of A. Thacker (1850), see [Dickson, Brown]. - Franz Vrabec, Aug 21 2005

G.f. A(x) satisfies: A(x) = x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^6 - Sum_{k>=2} k^4 * A(x^k). - Ilya Gutkovskiy, Mar 29 2020

Sum_{k=1..n} a(k) ~ n^6 / (5*Pi^2). - Amiram Eldar, Dec 03 2023